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Let x chairs and y tables were produced.
Number of chairs and tables cannot be negative.
Therefore, x, y ≥ 0
The given information can be tabulated as follows:
Time on machine A(hrs) Time on machine B (hrs)
Chairs 2 6
Tables 4 2
Availability 16 30
Therefore, the constraints are
2x+4y< 16
6x+ 2y < 30
Profit gained by the manufacturer from a chair and a table is Rs 3 and Rs 5 respectively.
Therefore, profit gained from x chairs and y tables is Rs 3x and Rs 5y.
Total profit Z = 3m + 5y which is to be maximized
Thus, the mathematical formulation of the given linear programmimg problem is
MaxZ= 3m + 5y
subject to
2x+4y< 16
6x+ 2y < 30
First we will convert inequations into equations as follows:
2x+4y ≥ O
Region represented by 2x+ 4y ≤ 16:
The line 2x+4y= 16 meets the coordinate axes at A1 and B1 (0, 4) respectively. By joining these points we obtain
the line 2x + 4y =16. Clearly (0,0) satisfies the 2x+ 4y ≤=16. So, the region which contains the origin represents the
solution set of the inequation 2x+ 4y ≤ 16.
Region represented by 6x+ 2y≤ 30:
The line 6x + 2y =30 meets the coordinate axes at Cl (5, 0) and DI (0, 15) respectively. By joining these points we obtain the line 6x + 2y . Clearly (0,0) satisfies the inequation 6x + 2y 30. So,the region which contains the origin represents the solution set of the inequation 6x + 2y ≤ 30.
Region represented by x ≥ 0 and y≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y≥ 0.
The feasible region determined by the system of constraints 2x+ 4y ≤ 16, 6x+ 2y≤ 30, x ≥ 0 and y≥ 0 are as follows.
The values of Z at these corner points are as follows
Corner point Z= 3x+ 5y
O 0
B1 20
E1 22.2
C1 15
The maximum value of Z is 22.2 which is attained at B1
(22/5, 9/5) Thus, the maximum profit is of Rs 22.20 obtained when units of chairs (22/5) and (9/5) units of tables are produced.
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